Print first N characters of a string (C)

Question

My objective is to print out the first N characters (the len variable) of a string using a simple for loop. However, the code doesn't seem to work. eg with the input printFrstChars("Example",3) the code should print out Exa , but it does not print out anything .

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

void printFrstChars (char inp[], int len)
{
    for(int i = 0; inp[i] != '\0' && i <= len; i++){
        printf("%s", inp[i]);

    }

}

int main ()
{   
    int len = 0;
    char inp[100];
    printf("Input string and length:\n");
    scanf("%99s %d", &inp, &len);
    printFrstChars(inp, len);
   

    return 0;
}

I went over the code multiple times and I (obviously) did not find the mistake that causes this. I would guess that it is either a mistake subtle enough to hide from my beginner eye, or the whole approach is wrong.

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

void printFrstChars (char inp[], int len)
{
    for(int i = 0; inp[i] != '\0' && i < len; i++){
        printf("%c", inp[i]);
    }
    printf("\n");
}

int main ()
{   
    int len = 0;
    char inp[100];
    printf("Input string:\n");
    scanf("%99s", inp);
    printf("Input length (99 or less):\n");
    scanf("%d", &len);
    printFrstChars(inp, len);
   

    return 0;
}

Separating inputs, scanning into inp instead of &inp , printing each char with %c format, and iterating while i < len instead (printing a new line after). My 2 cents.

---

Edit

Thanks to @AndrewHenle for pointing out the overflow. The %99s format specifier only reads max 99 chars (and the prompt states that too).

Answer 2

I think you need to remove the '99' from %99s where you are taking the string as an input. Just %s will work in the case, you are not entering a string with 'spaces'. Also, In case you want to print the N number of characters, you need to put len<N or len<=N-1. Just, put inp instead of &inp.